The differential equation of the family of ci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which sim

Vedclass · Accepted Answer

$y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}+1=0$

Vedclass · Answer

(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 = a^2$,or $x^2 + y^2 - 2ax = 0$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $a = x + y \frac{dy}{dx}$.Substituting the value of $a$ back into the original equation: $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \Rightarrow y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.However,for a general family of circles with center $(a, 0)$ and radius $r$,the equation is $(x-a)^2 + y^2 = r^2$. Differentiating twice eliminates two constants $a$ and $r$. The standard form for circles with center on the $x$-axis is $y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

The differential equation of the family of circles whose center lies on the $x$-axis is:

Similar Questions

If $\sqrt{y}=\cos ^{-1} x$,then it satisfies the differential equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=c$,where $c$ is equal to

Form the differential equation of the family of hyperbolas having foci on the $x$-axis and centre at the origin.

$y = ax + b$ is

The differential equation representing the family of circles having their centres on the $Y$-axis is (where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$):

If $m$ and $n$ are the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module